POLI 325 due 10/13/04
EXERCISES PERTAINING TO VOTING RULES: ANSWERS & DISCUSSION
1. Preference Profile 1
# of voters 4 4 2 9
1st pref. A B B C
2nd pref. B A D D
3rd pref. D D A A
4th pref. C C C B
Voting Rule Winning Candidate
Simple Plurality Voting A: 4 votes, B: 6, C: 9, D: 0, so C wins
Approval Voting (when each voter votes
for two candidates) A: 8 votes, B: 10, C: 9, D: 11, so D wins
Plurality Runoff (or IRV variant 1) C and B are the top two candidates
in the first round, B wins runoff
Alternative Vote (or IRV variant 2) Same as above, since only three candidates win votes in the first round, so B wins
Borda Point Voting Calculate the Borda scores: Borda Ordering
A = 4*4 + 4*3 + 2*2 + 9*2 = 16 + 12 + 4 + 18 = 50 A (winner)
B = 4*3 + 4*4 + 2*4 + 9*1 = 12 + 16 + 8 + 9 = 45 D
C = 4*1 + 4*1 + 2*1 + 9*4 = 4 + 4 + 2 + 36 = 46 C
D = 4*2 + 4*2 + 2*3 + 9*3 = 8 + 8 + 6 + 27 = 49 B
Note 1. I used, as is standard, the scores 4, 3, 2, 1 (in general, m, m-1, m-2, . . . , where m is the number of candidates) for 1st preferences, 2nd preferences, etc. However, any decreasing sequences of numbers at equal intervals apart (any “linear transformation”), e.g., 3, 2, 1, 0 or 10, 7, 4, 1, etc., produces the same Borda winner and Border ordering.
Note 2. Given the standard scoring system, each voter’s ballot awards a total of m + (m-1) + (m-2) . . . + 1 Borda points — in this case 10 + 3 + 2 + 1 = 10 points. Thus a total of 190 points are awarded. This provides a useful arithmetic check, i.e., 50 + 49 + 46 + 45 = 190.
Condorcet Voting Calculate each straight fight:
A beats B, 13 to 6 B beats D. 10 to 9
A beats C, 10 to 9 D beats A, 11 to 8
B beats C, 10 to 9 D beats C, 10 to 9
C is beaten by all other candidates, and is therefore the Condorcet loser. But A, B, and D are in a majority cycle, so there is no Condorcet winner and Condorcet Voting does not work.
Note. Given m candidates, there are “m choose 2” straight fights, i.e., m!/2(m-2)! = m(m-1)/2 straight fights (where n! = n factorial = n × (n-1) . . . × 2 × 1). Note that this number increases much faster than m does.
Here are two more voting systems:
Coombs Voting: proceed in the manner of the Alternative Vote (see p. 3 in Voting to Elect Several Candidates) but, instead of eliminating the remaining candidate with the fewest first preferences, eliminate the candidate with the most last preferences.
Sequential ( or “Knockout”) Voting: pair two candidates in a straight fight and eliminate the loser; pair the winner with a third candidate and eliminate the loser; proceed until every candidate but one has been eliminated; the last candidate standing is the winner.
Voting Rule Winning Candidate
Coombs Voting C has the most last preferences (10) and is eliminated; among A, B, and D, B has the most last preferences (9) and is eliminated; between A and D, A has the most last preferences (11), so D wins. Note that in the last round Coombs voting is the same as simple majority rule.
Note. We have seen that a Condorcet winner can lose under either variant of IRV, because the Condorcet winner may have the fewest first preferences (possibly none if m > 4). It is also true that the Condorcet winner may have the most last preferences, so the Condorcet winner can also lose under Coombs Voting. Here is an example:
3 2 2 2
A A B C
C B C B
B C A A
Sequential Voting (when the Paired votes (straight fights)
candidates are paired in A vs. B B is knocked out, A survives
alphabetical order) A vs. C C is knocked out, A survives
A vs. D A is knocked out, so D wins
Does the Sequential Voting winner change when the candidates are paired in other orders?
Let’s try reverse alphabetical D vs. C C is knocked out, D survives
order D vs. B D is knocked out, B survives
B vs. A B is knocked out, A wins
So the answer is yes.
We can observe the following additional points. A Condorcet loser can’t survive any vote and therefore can’t win under any voting order. So C can’t win in any event. In contrast, a Condorcet winner can never be knocked out and therefore must win under any voting order. But there is no Condorcet winner in Profile 1. Instead there are three candidates A, B, and D in a cycle. The last one to enter the voting order wins, because the winner of the first pairing of the three “cycling” candidates must subsequently be knocked out by the third.
2. You should have found that, under Preference Profile 1, both variants of IRV (equivalent to Plurality Runoff and Alternative Vote, respectively — see footnote 1 in Voting to Elect Several Candidates) give the same winner. Can you construct a preference profile in which the two variants produce different winners?
Number of voters: n1 > n2 > n3 > n4 where n1 < n2 + n3 + n4
and n3 + n4 > n2
1st pref. A B C D
2nd pref. ? C ? C
3rd pref. ? ? ? ?
4th pref. ? ? ? ?
IRV variant 1 (simultaneous elimination) C and D are eliminated after the first round so either A or B wins (depending on the lower preferences of the third and fourth groups of voters)
IRV variant 2 (sequential elimination) D is eliminated first, so n4 votes transfer to C, putting C above B, so B is eliminated next. The final runoff is between A and C, which and C wins.
3. Preference Profile 2
# of voters 4 2 2 2 4 1
1st pref. A A B B C C
2nd pref. B C A C B A
3rd pref. C B C A A B
Voting Rule Winning Candidate
Simple Plurality Voting A: 6 votes, B: 4, C: 5, so A wins
Approval Voting (when each voter votes A: 9 votes, B: 12, C: 9, so B wins
for two candidates)
Plurality Runoff (or IRV variant 1) Runoff between A and C, A wins
Alternative Vote (or IRV variant 2) With just 3 candidates, the two IRV variants are equivalent, so A wins
Borda Ordering
Borda A = 4*3+2*3+2*2+2*1+4*1+ 1*2 = 12+6+4+2+4+2 = 30 B (winner)
Point B = 4*2+2*1+2*3+2*3+4*2+1*1 = 8+2+6+6+8+1 = 31 A
Voting C = 4*1+2*2+2*1+2*2+4*3+1*3 = 4+4+2+4+12+3 = 29 C
Condorcet Voting Straight Fights Condorcet ordering
B beats A, 8 to 7 B (winner)
B beats C, 8 to 7 A
A beats C, 8 to 7 C
Coombs Voting A and C are tied for the most last preference but, whichever is eliminated first, B wins
Sequential Voting (when the A vs. B A knocked out, B survives
candidates are paired in B vs. C C knocked out, B wins
alphabetical order)
Does the Sequential Voting winner change when the candidates are paired in other orders?
No. As the Condorcet winner, B wins whatever the voting order. Note that Profile 2 is not “single peaked” and includes all possible orderings of three candidates including “inadmissible orderings” (as discussed on the top of p. 6 of in Voting to Elect a Single Candidate). Nevertheless, there is a Condorcet winner. Here is why. With respect to straight fights (and thus Condorcet and Sequential Voting), voters with directly opposed preference orderings “cancel each other out,” so deleting pairs of such voters from the preference profile doesn’t affect the outcome of any straight fight. You can check that deleting all such pairs reduces Profile 2 to a single voter who prefers B to A to C, whose preferences therefore wholly determine the Condorcet ordering (and guarantee that there is a Condorcet winner) and the winners under both Condorcet and Sequential Voting. In contrast, deleting (or adding) pairs of voters with directly opposed preferences may change the winning outcome under others voting rules (that are not based on straight fights).
4. The 15 members of a Congressional committee are choosing among five different proposed funding levels for a government program. Here are the proposals and the first preferences of all committee members.
Proposal 1 $10 million which is the first preference of 5 members
Proposal 2 $12 million which is the first preference of 2 members
Proposal 3 $18 million which is the first preference of 1 member
Proposal 4 $19 million which is the first preference of 4 members
Proposal 5 $21 million which is the first preference of 3 members
With respect to second and lower preferences, each committee member ranks proposals according to their proximity (or “closeness”) to his or her first preference. For example, the preference ordering of the member who most prefers Proposal 3 is this:
Proposal Proximity to first pref.
first pref. $18 $0
second pref. $19 $1
third pref. $21 $3
fourth pref. $12 $6
fifth pref. $10 $8
The Committee will adopt one proposal on the basis of Sequential Voting (which is more or less how Congressional bodies actually take votes). Can you determine which proposal will win in committee? Does the winning proposal depend on the order in which proposals may be paired for votes?
Hint: Is there a Condorcet winner among the five proposals? If so, is there some way by which we can quickly determine which proposal it is?
We know that, if there is a Condorcet winner, it wins under Sequential Voting regardless of the order in which proposals are voted on. Thus, the questions are whether there is a Condorcet winner and, if so, which proposal it is.
Given the preferences specified above, the voters can be ranked from the “smallest spenders” to the “biggest spenders.” In this ordering, the member who wants to spend $18 million occupies the median position, i.e., fewer than half of the members want to spend less and fewer than half the voters want to spend more. Therefore, the proposal that is the first preference of the median voter can beat any alternative proposal, i.e., it is the Condorcet winner. Thus the median voter rules. (If you follow Supreme Court decisions, think of Justice Sandra Day O’Connor.) So we know that, whatever other spending proposals (in addition to the $18 million proposal) are made and whatever order they are voted on, the proposal to spend $18 million must prevail. This result is known as the Median Voter Theorem and is originally due to Duncan Black (“On the Rationale of Group Decision-Making,” Journal of Political Economy, February 1948). We will see it again when we consider the logical of two-party competition.
5. For “Extra Credit”: Show by example that — as claimed on p. 5 of Voting to Elect Several Candidates — no PR apportionment formula can satisfy both Intra-Election Monotonicity. and Inter-Election Monotonicity. Hint: consider what any PR formula must do in the special case in which m = 1 (e.g., when applied in a single-member district).
Remember the definitions:
(a) Intra-Election Monotonicity. If party A gets a bigger vote share in an election than party B, A should get no fewer seats than B does.
(b) Inter-Election Monotonicity. If party A gets a bigger vote share in Election 2 than in Election 1 (the number of seats remaining constant), A should get no fewer seats in Election 2 than in Election 1.
Every “reasonable” apportionment formula meets condition (a). This means that every “reasonable” PR formula, when applied to a Single Member District, is equivalent to Simple Plurality Voting, i.e., the party with the most first preferences wins the single seat.
It is easy to devise a pair of single-winner elections in which Simple Plurality Rule violates condition (b).
Vote Share for => Party A Party B Party C
Election 1 40% 35% 25% By (a), A must win the seat.
Election 2 42% 45% 13% By (a) B must win the seat, but by (b) A can’t lose it.
So in general no “reasonable” PR apportionment formula can meet condition (b).